3.397 \(\int \cos ^2(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=131 \[ -\frac {a^3 \cos ^5(c+d x)}{5 d}+\frac {a^3 \cos ^3(c+d x)}{d}+\frac {3 a^3 \cos (c+d x)}{d}-\frac {a^3 \cot (c+d x)}{d}-\frac {3 a^3 \sin ^3(c+d x) \cos (c+d x)}{4 d}+\frac {11 a^3 \sin (c+d x) \cos (c+d x)}{8 d}-\frac {3 a^3 \tanh ^{-1}(\cos (c+d x))}{d}-\frac {3 a^3 x}{8} \]

[Out]

-3/8*a^3*x-3*a^3*arctanh(cos(d*x+c))/d+3*a^3*cos(d*x+c)/d+a^3*cos(d*x+c)^3/d-1/5*a^3*cos(d*x+c)^5/d-a^3*cot(d*
x+c)/d+11/8*a^3*cos(d*x+c)*sin(d*x+c)/d-3/4*a^3*cos(d*x+c)*sin(d*x+c)^3/d

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Rubi [A]  time = 0.20, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2872, 3770, 3767, 8, 2638, 2635, 2633} \[ -\frac {a^3 \cos ^5(c+d x)}{5 d}+\frac {a^3 \cos ^3(c+d x)}{d}+\frac {3 a^3 \cos (c+d x)}{d}-\frac {a^3 \cot (c+d x)}{d}-\frac {3 a^3 \sin ^3(c+d x) \cos (c+d x)}{4 d}+\frac {11 a^3 \sin (c+d x) \cos (c+d x)}{8 d}-\frac {3 a^3 \tanh ^{-1}(\cos (c+d x))}{d}-\frac {3 a^3 x}{8} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*Cot[c + d*x]^2*(a + a*Sin[c + d*x])^3,x]

[Out]

(-3*a^3*x)/8 - (3*a^3*ArcTanh[Cos[c + d*x]])/d + (3*a^3*Cos[c + d*x])/d + (a^3*Cos[c + d*x]^3)/d - (a^3*Cos[c
+ d*x]^5)/(5*d) - (a^3*Cot[c + d*x])/d + (11*a^3*Cos[c + d*x]*Sin[c + d*x])/(8*d) - (3*a^3*Cos[c + d*x]*Sin[c
+ d*x]^3)/(4*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2872

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Dist[1/a^p, Int[ExpandTrig[(d*sin[e + f*x])^n*(a - b*sin[e + f*x])^(p/2)*(a + b*sin[e + f*x]
)^(m + p/2), x], x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, n, p/2] && ((GtQ[m,
0] && GtQ[p, 0] && LtQ[-m - p, n, -1]) || (GtQ[m, 2] && LtQ[p, 0] && GtQ[m + p/2, 0]))

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \cos ^2(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^3 \, dx &=\frac {\int \left (a^7+3 a^7 \csc (c+d x)+a^7 \csc ^2(c+d x)-5 a^7 \sin (c+d x)-5 a^7 \sin ^2(c+d x)+a^7 \sin ^3(c+d x)+3 a^7 \sin ^4(c+d x)+a^7 \sin ^5(c+d x)\right ) \, dx}{a^4}\\ &=a^3 x+a^3 \int \csc ^2(c+d x) \, dx+a^3 \int \sin ^3(c+d x) \, dx+a^3 \int \sin ^5(c+d x) \, dx+\left (3 a^3\right ) \int \csc (c+d x) \, dx+\left (3 a^3\right ) \int \sin ^4(c+d x) \, dx-\left (5 a^3\right ) \int \sin (c+d x) \, dx-\left (5 a^3\right ) \int \sin ^2(c+d x) \, dx\\ &=a^3 x-\frac {3 a^3 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {5 a^3 \cos (c+d x)}{d}+\frac {5 a^3 \cos (c+d x) \sin (c+d x)}{2 d}-\frac {3 a^3 \cos (c+d x) \sin ^3(c+d x)}{4 d}+\frac {1}{4} \left (9 a^3\right ) \int \sin ^2(c+d x) \, dx-\frac {1}{2} \left (5 a^3\right ) \int 1 \, dx-\frac {a^3 \operatorname {Subst}(\int 1 \, dx,x,\cot (c+d x))}{d}-\frac {a^3 \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}-\frac {a^3 \operatorname {Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {3 a^3 x}{2}-\frac {3 a^3 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {3 a^3 \cos (c+d x)}{d}+\frac {a^3 \cos ^3(c+d x)}{d}-\frac {a^3 \cos ^5(c+d x)}{5 d}-\frac {a^3 \cot (c+d x)}{d}+\frac {11 a^3 \cos (c+d x) \sin (c+d x)}{8 d}-\frac {3 a^3 \cos (c+d x) \sin ^3(c+d x)}{4 d}+\frac {1}{8} \left (9 a^3\right ) \int 1 \, dx\\ &=-\frac {3 a^3 x}{8}-\frac {3 a^3 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {3 a^3 \cos (c+d x)}{d}+\frac {a^3 \cos ^3(c+d x)}{d}-\frac {a^3 \cos ^5(c+d x)}{5 d}-\frac {a^3 \cot (c+d x)}{d}+\frac {11 a^3 \cos (c+d x) \sin (c+d x)}{8 d}-\frac {3 a^3 \cos (c+d x) \sin ^3(c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]  time = 2.05, size = 148, normalized size = 1.13 \[ \frac {(a \sin (c+d x)+a)^3 \left (-60 (c+d x)+80 \sin (2 (c+d x))+15 \sin (4 (c+d x))+580 \cos (c+d x)+30 \cos (3 (c+d x))-2 \cos (5 (c+d x))+80 \tan \left (\frac {1}{2} (c+d x)\right )-80 \cot \left (\frac {1}{2} (c+d x)\right )+480 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-480 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{160 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^6} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*Cot[c + d*x]^2*(a + a*Sin[c + d*x])^3,x]

[Out]

((a + a*Sin[c + d*x])^3*(-60*(c + d*x) + 580*Cos[c + d*x] + 30*Cos[3*(c + d*x)] - 2*Cos[5*(c + d*x)] - 80*Cot[
(c + d*x)/2] - 480*Log[Cos[(c + d*x)/2]] + 480*Log[Sin[(c + d*x)/2]] + 80*Sin[2*(c + d*x)] + 15*Sin[4*(c + d*x
)] + 80*Tan[(c + d*x)/2]))/(160*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6)

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fricas [A]  time = 0.48, size = 147, normalized size = 1.12 \[ -\frac {30 \, a^{3} \cos \left (d x + c\right )^{5} - 5 \, a^{3} \cos \left (d x + c\right )^{3} + 60 \, a^{3} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 60 \, a^{3} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 15 \, a^{3} \cos \left (d x + c\right ) + {\left (8 \, a^{3} \cos \left (d x + c\right )^{5} - 40 \, a^{3} \cos \left (d x + c\right )^{3} + 15 \, a^{3} d x - 120 \, a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{40 \, d \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/40*(30*a^3*cos(d*x + c)^5 - 5*a^3*cos(d*x + c)^3 + 60*a^3*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 60*a^3
*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 15*a^3*cos(d*x + c) + (8*a^3*cos(d*x + c)^5 - 40*a^3*cos(d*x + c)
^3 + 15*a^3*d*x - 120*a^3*cos(d*x + c))*sin(d*x + c))/(d*sin(d*x + c))

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giac [A]  time = 0.27, size = 226, normalized size = 1.73 \[ -\frac {15 \, {\left (d x + c\right )} a^{3} - 120 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 20 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {20 \, {\left (6 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} + \frac {2 \, {\left (55 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 200 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 10 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 720 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 800 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 10 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 560 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 55 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 152 \, a^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{40 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/40*(15*(d*x + c)*a^3 - 120*a^3*log(abs(tan(1/2*d*x + 1/2*c))) - 20*a^3*tan(1/2*d*x + 1/2*c) + 20*(6*a^3*tan
(1/2*d*x + 1/2*c) + a^3)/tan(1/2*d*x + 1/2*c) + 2*(55*a^3*tan(1/2*d*x + 1/2*c)^9 - 200*a^3*tan(1/2*d*x + 1/2*c
)^8 - 10*a^3*tan(1/2*d*x + 1/2*c)^7 - 720*a^3*tan(1/2*d*x + 1/2*c)^6 - 800*a^3*tan(1/2*d*x + 1/2*c)^4 + 10*a^3
*tan(1/2*d*x + 1/2*c)^3 - 560*a^3*tan(1/2*d*x + 1/2*c)^2 - 55*a^3*tan(1/2*d*x + 1/2*c) - 152*a^3)/(tan(1/2*d*x
+ 1/2*c)^2 + 1)^5)/d

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maple [A]  time = 0.42, size = 152, normalized size = 1.16 \[ -\frac {a^{3} \left (\cos ^{5}\left (d x +c \right )\right )}{5 d}-\frac {a^{3} \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{4 d}-\frac {3 a^{3} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{8 d}-\frac {3 a^{3} x}{8}-\frac {3 a^{3} c}{8 d}+\frac {a^{3} \left (\cos ^{3}\left (d x +c \right )\right )}{d}+\frac {3 a^{3} \cos \left (d x +c \right )}{d}+\frac {3 a^{3} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}-\frac {a^{3} \left (\cos ^{5}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*csc(d*x+c)^2*(a+a*sin(d*x+c))^3,x)

[Out]

-1/5*a^3*cos(d*x+c)^5/d-1/4*a^3*cos(d*x+c)^3*sin(d*x+c)/d-3/8*a^3*cos(d*x+c)*sin(d*x+c)/d-3/8*a^3*x-3/8/d*a^3*
c+a^3*cos(d*x+c)^3/d+3*a^3*cos(d*x+c)/d+3/d*a^3*ln(csc(d*x+c)-cot(d*x+c))-1/d*a^3/sin(d*x+c)*cos(d*x+c)^5

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maxima [A]  time = 0.45, size = 141, normalized size = 1.08 \[ -\frac {32 \, a^{3} \cos \left (d x + c\right )^{5} - 80 \, {\left (2 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right ) - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{3} - 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} + 80 \, {\left (3 \, d x + 3 \, c + \frac {3 \, \tan \left (d x + c\right )^{2} + 2}{\tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} a^{3}}{160 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/160*(32*a^3*cos(d*x + c)^5 - 80*(2*cos(d*x + c)^3 + 6*cos(d*x + c) - 3*log(cos(d*x + c) + 1) + 3*log(cos(d*
x + c) - 1))*a^3 - 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a^3 + 80*(3*d*x + 3*c + (3*tan(d
*x + c)^2 + 2)/(tan(d*x + c)^3 + tan(d*x + c)))*a^3)/d

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mupad [B]  time = 8.81, size = 356, normalized size = 2.72 \[ \frac {3\,a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {-\frac {13\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{2}+20\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+72\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-10\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+80\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-11\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+56\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}+\frac {76\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{5}-a^3}{d\,\left (2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}+\frac {3\,a^3\,\mathrm {atan}\left (\frac {9\,a^6}{16\,\left (\frac {9\,a^6}{2}+\frac {9\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}\right )}-\frac {9\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,\left (\frac {9\,a^6}{2}+\frac {9\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}\right )}\right )}{4\,d}+\frac {a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^4*(a + a*sin(c + d*x))^3)/sin(c + d*x)^2,x)

[Out]

(3*a^3*log(tan(c/2 + (d*x)/2)))/d + ((a^3*tan(c/2 + (d*x)/2)^2)/2 + 56*a^3*tan(c/2 + (d*x)/2)^3 - 11*a^3*tan(c
/2 + (d*x)/2)^4 + 80*a^3*tan(c/2 + (d*x)/2)^5 - 10*a^3*tan(c/2 + (d*x)/2)^6 + 72*a^3*tan(c/2 + (d*x)/2)^7 - 4*
a^3*tan(c/2 + (d*x)/2)^8 + 20*a^3*tan(c/2 + (d*x)/2)^9 - (13*a^3*tan(c/2 + (d*x)/2)^10)/2 - a^3 + (76*a^3*tan(
c/2 + (d*x)/2))/5)/(d*(2*tan(c/2 + (d*x)/2) + 10*tan(c/2 + (d*x)/2)^3 + 20*tan(c/2 + (d*x)/2)^5 + 20*tan(c/2 +
(d*x)/2)^7 + 10*tan(c/2 + (d*x)/2)^9 + 2*tan(c/2 + (d*x)/2)^11)) + (3*a^3*atan((9*a^6)/(16*((9*a^6)/2 + (9*a^
6*tan(c/2 + (d*x)/2))/16)) - (9*a^6*tan(c/2 + (d*x)/2))/(2*((9*a^6)/2 + (9*a^6*tan(c/2 + (d*x)/2))/16))))/(4*d
) + (a^3*tan(c/2 + (d*x)/2))/(2*d)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*csc(d*x+c)**2*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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